Find the minimum kinetic energy (in joule) with which a particle of mass `2kg` should be projected from ground so that it crosses a cylindrical drum (placed on ground) of radius 1 meter and fall on the other side following a parabolic path. [Tak `g=10m//s^(2)` and `sqrt(2)=1.414`]

Explanation

To determine the minimum kinetic energy with which a particle of mass 2 kg should be projected to cross a cylindrical drum of radius 1 meter, we can follow these steps:

### Step 1: Understand the Problem
We need to find the minimum kinetic energy required for a particle to be projected such that it crosses a cylindrical drum of radius 1 meter and falls on the other side.

### Step 2: Determine the Maximum Height
The particle must reach a height equal to the radius of the drum (1 meter) to clear it. The maximum height \( h \) reached in projectile motion can be expressed as:
\[ h = \frac{v^2 \sin^2 \theta}{2g} \]
where \( v \) is the initial velocity, \( \theta \) is the angle of projection, and \( g \) is the acceleration due to gravity (10 m/s²).

### Step 3: Set the Angle for Maximum Range
To achieve the maximum range, the angle of projection should be 45 degrees. Therefore, \( \sin 2\theta = \sin 90^\circ = 1 \).

### Step 4: Calculate the Required Range
To clear the drum, the total horizontal distance (range) the particle must cover is:
\[ \text{Range} = 2 \times \text{radius} = 2 \times 1 = 2 \text{ meters} \]

### Step 5: Use the Range Formula
The formula for the range \( R \) in projectile motion is given by:
\[ R = \frac{v^2 \sin 2\theta}{g} \]
Substituting \( R = 2 \) meters, \( g = 10 \) m/s², and \( \sin 2\theta = 1 \):
\[ 2 = \frac{v^2 \cdot 1}{10} \]
From this, we can solve for \( v^2 \):
\[ v^2 = 2 \times 10 = 20 \]

### Step 6: Calculate the Initial Velocity
Taking the square root to find \( v \):
\[ v = \sqrt{20} = 4.472 \text{ m/s} \]

### Step 7: Calculate the Minimum Kinetic Energy
The kinetic energy \( KE \) is given by:
\[ KE = \frac{1}{2} mv^2 \]
Substituting \( m = 2 \) kg and \( v^2 = 20 \):
\[ KE = \frac{1}{2} \times 2 \times 20 = 20 \text{ joules} \]

### Final Answer
Thus, the minimum kinetic energy required is:
\[ \boxed{20 \text{ joules}} \]